{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

equations

equations - Solving Equations Introductory example Solve...

This preview shows pages 1–3. Sign up to view the full content.

Solving Equations Introductory example . Solve the equation = x ( ) + x 2 15 ------- (i). We need to find all the real numbers x such the statement = x ( ) + x 2 15 is true. Putting the problem into words, we need to find a number such that the product of this number with a second number, that is obtained by adding 2 to the first number, is 15. Thinking just in terms of integers (whole numbers) we can quickly find the solution = x 3 by a "trial and error" approach. However, there is another solution. Rewriting the equation in the form = + x 2 2 x 15 by expanding the left side using distributivity of multiplication over addition. Then we can rewrite the equation in the form = + - x 2 2 x 15 0. so that the left side can be factored to give = ( ) - x 3 ( ) + x 5 0 ------- (ii). At this stage we can use the fact that a product of two real numbers a and b is equal to zero exactly when either one or other of the two numbers a and b is equal to zero, that is, = a b 0 exactly when = a 0 or = b 0. ________________ This is equivalent to the following two facts taken together. Multiplying any real number by 0 gives 0, that is, = 0 . a 0 for any real number a . Multiplying two non-zero numbers together gives a product that is not zero, that is, if a 0 and b 0, then a b 0. We apply this "zero product principle" to equation (ii) to see that = ( ) - x 3 ( ) + x 5 0 exactly when either = - x 3 0 or = + x 5 0, which means the same as saying = x 3 or = x - 5. Terminology used in connection with solving equations . If we introduce the concept of the solution set of an equation as the set of all solutions, then the solution set of equation (i) is { , - 5 3}, that is, the solution set is { x | x is a real number and = x ( ) + x 2 15 } = { , - 5 3}. In general, two equations are said to be equivalent exactly when they have the same solution set. The process of solving an equation (usually) involves writing down a sequence of equivalent equations, ending up with an equation from which the solutions can readily be obtained. The logical connective <=> which means "is equivalent to" may be used here. Thus the solution of equation (i) can be written down as follows. = x ( ) + x 2 15 <=> = + x 2 2 x 15 <=> = - - x 2 2 x 15 0 <=> = ( ) - x 3 ( ) + x 5 0 <=> = x 3 or = x - 5. _______ Each occurrence of <=> can be read as either "is equivalent to" or "which is equivalent to". One can then finish by stating that the solution set of the original equation is { , 3 - 5}. The symbol <=> is often omitted, however.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Processes that produce an equivalent equation when applied to an equation involving a single variable or "unknown" x . Adding a constant or an algebraic expression involving the variable x to both sides of an equation provided that the expression added can be evaluated for all real numbers. ( More generally, it is sufficient for the expression that is to be added to both sides to have a real number value for each of the solutions of the equation, but, since these are not known initially, it is safer to make sure that the expression to be added can be evaluated for all real numbers numbers.) Note that subtracting a positive number a from each side of an equation is the same as adding the negative number - a to each side.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern